DC homopolar motor/generator

ABSTRACT

In accordance with the teachings described herein, a DC homopolar machine is provided. The DC homopolar machine may include a stator having a stator magnetic core and a permanent magnet, a rotor magnetic core supported within the stator to rotate relative to the stator, and non-magnetic material within a gap between the rotor magnetic core and the stator magnetic core. The stator, rotor and non-magnetic material form a magnetic circuit having a total reluctance, the total reluctance of the magnetic circuit being provided by a first reluctance of the permanent magnet and a second reluctance of the gap of non-magnetic material, with the first reluctance being substantially equal to the second reluctance.

FIELD

The technology described in this patent document relates generally to electric motors and generators. More particularly, a DC homopolar machine is described that is particularly useful for applications requiring a high energy efficiency.

BACKGROUND

The earliest electromotive machine producing rotary shaft power was the homopolar motor/generator. Conversion of electric power into mechanical shaft power was first demonstrated by Michael Faraday in the early 19th century. Variously known as a “homopolar,” “unipolar” or “monopolar” machine, this unique energy conversion device represents a true DC machine requiring no commutator or other switching methods for converting direct current into alternating current as required by conventional so-called “DC” machines. The standard homopolar motor/generator is a single-turn machine requiring very high current at just a few volts. This peculiar characteristic of a homopolar machine renders it impractical for most commercial applications.

Creation of the magnetic field in a DC homopolar machine is readily facilitated by permanent magnet (PM) material. Historically, PM-based homopolar machine design has focused on creating the highest possible flux content from a given quantity of PM material. This design philosophy derives from the understanding that torque is directly proportional to total magnetic flux. Accordingly, the rotor-stator gap is typically kept as small as practicable in order to increase total flux content. Typical machines consequently embody a relatively small rotor-stator gap which in turn restricts the amount of copper available for current conduction through the rotor-stator gap region. Excessive heat production and low efficiency typically ensue as a result of reduced copper volume.

SUMMARY

In accordance with the teachings described herein, a DC homopolar machine is provided. The DC homopolar machine may include a stator having a stator magnetic core and a permanent magnet, a rotor magnetic core supported within the stator to rotate relative to the stator, and non-magnetic material within a gap between the rotor magnetic core and the stator magnetic core. The stator, rotor and non-magnetic material form a magnetic circuit having a total reluctance. The total reluctance of the magnetic circuit is provided by a first reluctance of the permanent magnet and a second reluctance of the gap of non-magnetic material, with the first reluctance being substantially equal to the second reluctance.

BRIEF DESCRIPTION OF THE DRAWINGS

FIG. 1 depicts a simple magnetic circuit that includes an ideal iron core of infinite permeability interrupted with two gaps.

FIG. 2 depicts a cross-sectional view of an example DC homopolar motor/generator.

FIG. 3 depicts an example external view of the DC homopolor motor/generator shown in FIG. 2.

DETAILED DESCRIPTION

FIG. 1 shows a simple magnetic circuit 10 that includes an ideal iron core 12 of infinite permeability interrupted with two gaps 14, 16. One gap 14 is filled with PM material and the other gap 16 is filled at least partial with copper conductors. Relative permeability of the PM gap 14 is nearly twice that of free space or air. The PM material 14 provides magnetomotive force (mmf) for driving flux through both its own reluctance and the reluctance of the other gap 16. The latter gap 16 is referred to herein as the “air gap” even though in an actual machine application it may be mostly filled with copper conductors. Copper has a relative permeability nearly that of air. Permanent magnets are rated for their “closed circuit” flux density which falls off when an air gap is added to the circuit due to the resulting rise of overall circuit reluctance. Thus, the actual gap flux density will always be less than the rated PM value depending on the magnitude of the additional air gap length.

Referring to FIG. 1, let the fundamental (mmf)_(cc) arising from the PM material be represented as: $\begin{matrix} {{({mmf})_{cc} = {({ni})_{cc} = \frac{B_{R}l_{m}}{\mu_{r}\mu_{o}}}};} & {{Eq}.\quad 1} \end{matrix}$ where:

-   -   (ni_(cc))=equivalent “closed circuit” amp-turns or mmf of PM         material that drives flux through the circuit;     -   B_(R)=so-called “Residual Induction”, the rated flux density of         PM material in a closed circuit;     -   l_(m)=gap length occupied by the PM material;     -   μ_(r)=relative magnetic permeability of PM material, where a         value of “unity” represents free space (air); and     -   μ_(o)=absolute permeability of free space (air essentially).         The magnetic circuit equation, similar in form to Ohm's Law, is:         (ni)_(cc)=φ         _(tot);  Eq. 2         where:     -   φ=circuit flux which has the same value at all points in the         circuit, thus no subscript; and     -   _(tot)=total circuit reluctance including both the air gap and         the gap occupied by PM material.         Total circuit reluctance         _(tot) is the summation of reluctance contributed by each gap,         _(m) and         _(g) where: $\begin{matrix}         {{R_{tot} = {\left( {R_{m} + R_{g}} \right) + \left( {\frac{l_{m}}{\mu_{r}\mu_{o}A_{m}} + \frac{l_{g}}{\mu_{o}A_{g}}} \right)}};} & {{Eq}.\quad 3}         \end{matrix}$         where:     -   _(m)=reluctance of gap occupied by PM material;     -   _(g)=reluctance of air gap (occupied by copper conductors);     -   A_(m)=gap area filled with PM material;     -   l_(g)=length of air gap; and     -   A_(g)=area of air gap.         The first term in Eq. 3 represents reluctance of the gap 14         holding PM material; the second term is the reluctance of the         air gap 16 containing the copper conductors.         Substituting Eq. 3 into Eq. 2: $\begin{matrix}         {({ni})_{cc} = {\phi\left( {\frac{l_{m}}{\mu_{r}\mu_{o}A_{m}} + \frac{l_{g}}{\mu_{o}A_{g}}} \right)}} & {{Eq}.\quad 4}         \end{matrix}$         Substituting Eq. 1 into Eq. 4: $\begin{matrix}         {\frac{B_{R}l_{m}}{\mu_{r}\mu_{o}} = {\phi\left( {\frac{l_{m}}{\mu_{r}\mu_{o}A_{m}} + \frac{l_{g}}{\mu_{o}A_{g}}} \right)}} & {{Eq}.\quad 5}         \end{matrix}$         The total circuit flux φ is the same everywhere, thus:         φ=A _(m) B _(m) =A _(g) B _(g) =A _(core) B _(sat);  Eq. 6         where:     -   B_(m)=flux density of PM gap under non-closed circuit         conditions;     -   B_(g)=flux density of air gap;     -   A_(core)=area of the circuit iron core connecting the two gaps;         and     -   B_(sat)=saturation flux density of the interconnecting core.

Core area A_(core) is the minimum value required to carry flux φ at the maximum saturation flux density in order to reduce circuit weight and achieve optimum core material (iron) utilization.

FIG. 2 depicts a cross-sectional view of an example DC homopolar machine 20. The example machine 20 depicted in FIG. 2 includes physical features that correspond to the magnetic circuit 10 depicted in FIG. 1. Specifically, the example homopolar machine includes a stator having a stator iron core 24 and a permanent magnet 22, a rotor magnetic core 28 supported within the stator 22, 24 to rotate relative to the stator, and non-magnetic material within a rotor-stator gap 26 between the rotor magnetic core 28 and the stator magnetic core 24. Also illustrated is a shaft 34 connected to the rotor 28. The non-magnetic material within the rotor-stator gap 26 depicted in this example includes a rotor copper conductor 30 and a stator copper conductor 32. The non-magnetic material in the rotor-stator gap 26 also includes at least a small gap filled with air, such that the rotor 28 may rotate relative to the stator 24. It should be understood, however, that in other examples the non-magnetic material within the rotor-stator gap 26 may include a larger gap of air, or may be entirely air without any copper conductors. In addition, other examples may include non-magnetic material within the rotor-stator gap 26 other than copper or air. FIG. 3 depicts an example external view of the DC homopolor motor/generator shown in FIG. 2.

As described in more detail with reference to the equations provided below, the stator 22, 24, rotor 28 and rotor-stator gap 26 form a magnetic circuit. The total reluctance of the magnetic circuit is provided by a first reluctance of the permanent magnet 22 and a second reluctance of the rotor-stator gap 26, with the first reluctance being substantially equal to the second reluctance. It should be understood that the term “substantially equal” is used herein to equate things that are either exactly equal or about equal.

Torque Equation Derivation:

The following equations (Eq. 7-Eq. 12) may be used to express the total machine torque for the example motor/generator depicted in FIGS. 2 and 3. Let motor torque T be given as: T=Fr _(g)=(IB _(g) w _(g))r _(g);  Eq. 7 where:

-   -   F=force developed on rotor conductors;     -   I=total electric current flowing through all the rotor         conductors;

B_(g)=rotor-stator gap magnetic flux density;

-   -   w_(g)=length of rotor conductors immersed in the magnetic field;         and     -   r_(g)=rotational radius of rotor conductors.         Let gap area A_(g) be expressed as the length of the gap         circumference 2πr_(g) times axial length w_(g):         A _(g)=(2πr _(g))w _(g)  Eq. 8         Solving Eq. 8 for w_(g) and substituting into Eq. 7:         $\begin{matrix}         {T = {{\left( \frac{{IB}_{g}A_{g}}{2\quad\pi\quad r_{g}} \right)r_{g}} = {\frac{1}{2\quad\pi}{IB}_{g}A_{g}}}} & {{Eq}.\quad 9}         \end{matrix}$         Inserting Eq. 6 into Eq. 9: $\begin{matrix}         {T = {\frac{1}{2\pi}I\quad\phi}} & {{Eq}.\quad 10}         \end{matrix}$

Total machine torque T is revealed by Eq. 10 to be a function of two fundamental machine properties, current capacity I, and machine flux capacity φ. Practical limitation of I is dictated by copper content. Maximum flux φ is governed by the iron content. The quantity of copper constrains current to within the thermal limit. Iron quantity limits flux to the saturation level. Rewriting Eq. 5 while using Eq. 6 for flux φ: $\begin{matrix} {{B_{R}l_{m}} = {A_{g}{B_{g}\left( {\frac{l_{m}}{A_{m}} + \frac{\mu_{r}l_{g}}{A_{g}}} \right)}}} & {{Eq}.\quad 11} \end{matrix}$ Solving Eq. 11 for rotor-stator gap flux B_(g): $\begin{matrix} {B_{g} = \frac{B_{R}l_{m}}{A_{g}\left( {\frac{l_{m}}{A_{m}} + \frac{\mu_{r}l_{g}}{A_{g}}} \right)}} & {{Eq}.\quad 12} \end{matrix}$ Machine Flux as Function of Rotor-Stator Gap:

If μ_(r)=1 and A_(m)=A_(g), then Eq. 12 reduces to: $\begin{matrix} {B_{g} = {{B_{R}\left( \frac{l_{m}}{l_{m} + l_{g}} \right)} = {\frac{B_{R}}{\left( {1 + \frac{l_{g}}{l_{m}}} \right)}\quad\left( {{{{at}\quad A_{m}} = A_{g}},{\mu_{r} = 1}} \right)}}} & {{Eq}.\quad 13} \end{matrix}$ From Eq. 13 it may be shown that gap flux density B_(g) declines from B_(g)=B_(R) at l_(g)=0 to B_(g)=½B_(R) at l_(g)=l_(m). Because torque is a function of total machine flux φ according to Eq. 10, not gap flux density B_(g), it is more instructive to rewrite Eq. 12 as: $\begin{matrix} {{A_{g}B_{g}} = {\phi = \frac{B_{R}l_{m}}{\left( {\frac{l_{m}}{A_{m}} + \frac{\mu_{r}l_{g}}{A_{g}}} \right)}}} & {{{Eq}.\quad 13}A} \end{matrix}$ Clearly, total machine flux diminishes as the rotor-stator gap length l_(g) increases, which is why conventional machine design typically attempts to keep the l_(g) as small as possible. Optimal Gap for Maximum Torque at Given Efficiency:

Solving Eq. 11 for A_(g) B_(g) and inserting into Eq. 10 results in: $\begin{matrix} {T = {\frac{1}{2\quad\pi}I\quad\frac{B_{R}l_{m}}{\left( {\frac{l_{m}}{A_{m}} + \frac{\mu_{r}l_{g}}{A_{g}}} \right)}}} & {{Eq}.\quad 14} \end{matrix}$

As shown by Eq. 14, maximum torque occurs at l_(g)=0 because this condition would give maximum gap flux density B_(g). In practice, however, at l_(g)=0 there would be no space available for copper, resulting in an infinitely high electrical resistance and zero machine efficiency. Conversely, an excessively large gap l_(g), while lowering electrical resistance, would also reduce the flux density, and, consequently, torque would drop at a given current I. Therefore, at a specified efficiency there is an optimum value of gap length l_(g) that produces peak torque.

To find the optimum rotor-stator gap length l_(g), begin by expressing motor efficiency as the ratio of mechanical power output to electrical power input: $\begin{matrix} {{E_{ff} = {\frac{P_{Shaft}}{P_{electric}} = {\frac{P_{S}}{\left( {P_{S} + P_{\Omega}} \right)} = \frac{1}{\left( {1 + \frac{P_{\Omega}}{P_{S}}} \right)}}}};} & {{Eq}.\quad 15} \end{matrix}$ where:

-   -   P_(S)=shaft power; and     -   P_(Ω)=resistive loss due to copper resistance

The importance of keeping resistive losses to a minimum to obtain good efficiency is shown by Eq. 15. The ratio P_(Ω)/P_(S) may be evaluated by first deriving the dissipative loss power P_(Ω): P _(Ω) =I ² R;  Eq. 16 where:

-   -   R=total rotor/stator resistance; and     -   I=total rotor/stator current

Machine resistance R is calculated using the dimensional information from FIG. 2: $\begin{matrix} {{R = \frac{2\quad\sigma\quad k_{w}w_{g}}{k_{F}A_{copper}}};} & {{Eq}.\quad 17} \end{matrix}$ where:

-   -   “2”=multiplier to account for both rotor and stator conductors         which are assumed to be identical in cross-sectional area and         length;     -   σ=copper resistivity;     -   k_(w)=proportional multiplier that determines total conductor         length as function of w_(g);     -   w_(g)=axial length of conductor immersed in the magnetic field         as previously defined for Eq. 7;     -   k_(F)=copper fill-factor applied to copper area A_(copper); and     -   A_(copper)=total area available for copper.         Solving Eq. 8 for w_(g) and inserting into Eq. 17:         $\begin{matrix}         {R = {\frac{2\sigma\quad k_{w}A_{g}}{2\pi\quad r_{g}k_{F}A_{copper}} = \frac{\sigma\quad k_{w}A_{g}}{\pi\quad r_{g}k_{F}A_{copper}}}} & {{Eq}.\quad 18}         \end{matrix}$         Let copper area A_(copper) be expressed as: $\begin{matrix}         {{A_{copper} = {{2{\quad\quad}\pi\quad{r_{g}\left( \frac{l_{g}}{2} \right)}} = {\pi\quad r_{g}l_{g}}}};} & {{Eq}.\quad 19}         \end{matrix}$         where l_(g)/2 accounts for half the radial rotor-stator gap         length allocated equally between rotor copper and stator copper         since rotor and stator conductors are connected in series.         Inserting Eq. 19 into Eq. 18: $\begin{matrix}         {R = {\frac{\sigma\quad k_{w}A_{g}}{\pi\quad r_{g}k_{F}\pi\quad r_{g}l_{g}} = \frac{\sigma\quad k_{w}A_{g}}{\pi^{2}k_{F}r_{g}^{2}l_{g}}}} & {{Eq}.\quad 20}         \end{matrix}$         Substituting Eq. 20 into Eq. 16: $\begin{matrix}         {P_{\Omega} = {I^{2}\frac{\sigma\quad k_{w}A_{g}}{\pi^{2}k_{F}r_{g}^{2}l_{g}}}} & {{Eq}.\quad 21}         \end{matrix}$         Let shaft power P_(S) be the product of torque T and angular         shaft frequency ω using Eq. 14 for T: $\begin{matrix}         {P_{S} = {{T\quad\omega} = {\frac{1}{2\quad\pi}I\quad\frac{\omega\quad B_{R}l_{m}}{\left( {\frac{l_{m}}{A_{m}} + \frac{\mu_{r}l_{g}}{A_{g}}} \right)}}}} & {{Eq}.\quad 22}         \end{matrix}$         Combining Eq. 21 and Eq. 22: $\begin{matrix}         \begin{matrix}         {\frac{P_{\Omega}}{P_{S}} = {\left( \frac{I^{2}\sigma\quad k_{w}A_{g}}{\pi^{2}k_{F}r_{g}^{2}l_{g}} \right)\left\lbrack \frac{2{\pi\left( {\frac{l_{m}}{A_{m}} + \frac{\mu_{r}l_{g}}{A_{g}}} \right)}}{I\quad\omega\quad B_{R}l_{m}} \right\rbrack}} \\         {= {I\quad\frac{2\quad\sigma\quad{k_{w}\left( {\frac{l_{m}A_{g}}{l_{g}A_{m}} + \mu_{r}} \right)}}{\pi\quad k_{F}r_{g}^{2}\omega\quad B_{R}l_{m}}}}         \end{matrix} & {{Eq}.\quad 23}         \end{matrix}$         Solving Eq. 23 for current I: $\begin{matrix}         {I = \frac{\pi\quad{k_{F}\left( {P_{\Omega}/P_{S}} \right)}B_{R}\omega\quad r_{g}^{2}l_{m}}{2\quad\sigma\quad{k_{w}\left( {\frac{l_{m}A_{g}}{l_{g}A_{m}} + \mu_{r}} \right)}}} & {{Eq}.\quad 24}         \end{matrix}$         Rewriting Eq. 14: $\begin{matrix}         {T = {I\quad\frac{B_{R}l_{m}A_{g}}{2\quad\pi\quad{l_{g}\left( {\frac{l_{m}A_{g}}{l_{g}A_{m}} + \mu_{r}} \right)}}}} & {{Eq}.\quad 25}         \end{matrix}$         Substituting Eq. 24 into Eq. 25: $\begin{matrix}         \begin{matrix}         {T = \frac{\pi\quad{k_{F}\left( {P_{\Omega}/P_{S}} \right)}B_{R}^{2}\omega\quad r_{g}^{2}l_{m}^{2}A_{g}}{4\sigma\quad k_{w}\pi\quad{l_{g}\left( {\frac{l_{m}A_{g}}{l_{g}A_{m}} + \mu_{r}} \right)}^{2}}} \\         {= {\frac{{k_{F}\left( {P_{\Omega}/P_{S}} \right)}B_{R}^{2}\omega\quad r_{g}^{2}}{4\sigma\quad k_{w}}\left\lbrack \frac{l_{m}^{2}A_{g}}{{l_{g}\left( {\frac{l_{m}A_{g}}{l_{g}A_{m}} + \mu_{r}} \right)}^{2}} \right\rbrack}}         \end{matrix} & {{Eq}.\quad 26}         \end{matrix}$         The variable configuration parameters are contained within the         square brackets of Eq. 26 which may be simplified to give:         $\begin{matrix}         {\left\lbrack \frac{l_{m}^{2}A_{g}}{{l_{g}\left( {\frac{l_{m}A_{g}}{l_{g}A_{m}} + \mu_{r}} \right)}^{2}} \right\rbrack = \frac{A_{g}l_{g}}{\left( {\frac{A_{g}}{A_{m}} + {\mu_{r}\frac{l_{g}}{l_{m}}}} \right)^{2}}} & {{Eq}.\quad 27}         \end{matrix}$

Examination of Eq. 27 shows there is an optimal value of either A_(g) or l_(g) that gives a maximum value of the quantity within the square brackets [ ]. The optimal value of l_(g) is found by differentiating the entire bracketed quantity [ ] with respect to l_(g) and setting the result equal to zero so that: $\begin{matrix} {{\frac{\mathbb{d}}{\mathbb{d}l_{g}}{\lbrack\rbrack}} = {{\frac{\mathbb{d}}{\mathbb{d}l_{g}}\frac{A_{g}l_{g}}{\left( {\frac{A_{g}}{A_{m}} + {\mu_{r}\frac{l_{g}}{l_{m}}}} \right)^{2}}} = 0}} & {{Eq}.\quad 28} \end{matrix}$ $\begin{matrix} {{\frac{\mathbb{d}}{\mathbb{d}l_{g}}{\lbrack\rbrack}} = {\left\lbrack {\frac{{- 2}\left( \frac{\mu_{r}}{l_{m}} \right)l_{g}}{\left( {\frac{A_{g}}{A_{m}} + {\mu_{r}\frac{l_{g}}{l_{m}}}} \right)^{3}} + \frac{1}{\left( {\frac{A_{g}}{A_{m}} + {\mu_{r}\frac{l_{g}}{l_{m}}}} \right)^{2}}} \right\rbrack = 0}} & {{Eq}.\quad 29} \end{matrix}$ $\begin{matrix} {{2\mu_{r}\frac{l_{g}}{l_{m}}} = \left( {\frac{A_{g}}{A_{m}} + {\mu_{r}\frac{l_{g}}{l_{m}}}} \right)} & {{Eq}.\quad 30} \end{matrix}$ $\begin{matrix} {\left( {\mu_{r}\frac{l_{g}}{l_{m}}} \right)_{opt} = \frac{A_{g}}{A_{m}}} & {{Eq}.\quad 31} \end{matrix}$ Rearranging the terms in Eq. 31: $\begin{matrix} {\left( \frac{l_{g}}{A_{g}} \right)_{opt} = {\frac{1}{\mu_{r}}\left( \frac{l_{m}}{A_{m}} \right)}} & {{Eq}.\quad 32} \end{matrix}$ Or: $\begin{matrix} {\left( {\mu_{r}\frac{l_{g}}{A_{g}}} \right)_{opt} = \frac{l_{m}}{A_{m}}} & {{{Eq}.\quad 32}A} \end{matrix}$

Thus, the optimal shape of the rotor-stator gap, that is, the proportional relationship of rotor-stator gap length l_(g) to rotor-stator gap area A_(g), is the same as the shape of the PM gap, except as modified by the multiplier (1/μ_(r)), where μ_(r) is the relative permeability of the PM material. Regardless of the absolute sizes of the rotor-stator gap and PM cavity volumes, their shape relative to each other is invariant as governed by Eq. 32. Furthermore, they may assume any imaginable shape, but nevertheless remain proportional to one another according to the dictates of Eq. 32. Optimal rotor-stator gap length is found by solving Eq. 32 for (l_(g))_(opt): $\begin{matrix} {\left( l_{g} \right)_{opt} = {\frac{1}{\mu_{r}}{l_{m}\left( \frac{A_{g}}{A_{m}} \right)}}} & {{Eq}.\quad 33} \end{matrix}$ Torque at Optimized Rotor-Stator Gap:

Torque production under the optimized condition stipulated by Eq. 33 is found by substituting Eq. 31 into Eq. 27: $\begin{matrix} \begin{matrix} {{\lbrack\rbrack}_{opt} = \frac{{A_{g}\left( l_{g} \right)}_{opt}}{\left( {\frac{A_{g}}{A_{m}} + \frac{A_{g}}{A_{m}}} \right)^{2}}} \\ {= \frac{{A_{g}\left( l_{g} \right)}_{opt}}{4\quad\frac{A_{g}^{2}}{A_{m}^{2}}}} \\ {= \frac{{A_{m}^{2}\left( l_{g} \right)}_{opt}}{4A_{g}}} \end{matrix} & {{Eq}.\quad 34} \end{matrix}$ Substituting Eq. 33 into Eq. 34: $\begin{matrix} \begin{matrix} {{\lbrack\rbrack}_{opt} = \frac{{A_{m}^{2}\left( l_{g} \right)}_{opt}}{4A_{g}}} \\ {= \frac{A_{m}^{2}l_{m}A_{g}}{4A_{g}\mu_{r}A_{m}}} \\ {= {\frac{1}{4}\frac{A_{m}l_{m}}{\mu_{r}}}} \end{matrix} & {{Eq}.\quad 35} \end{matrix}$ Substituting Eq. 35 into Eq. 26: $\begin{matrix} \begin{matrix} {T = {\frac{{k_{F}\left( {P_{\Omega}/P_{S}} \right)}B_{R}^{2}\omega\quad r_{g}^{2}}{16\sigma\quad k_{w}\mu_{r}}A_{m}l_{m}}} \\ {= {\frac{{k_{F}\left( {P_{\Omega}/P_{S}} \right)}\omega\quad r_{g}^{2}}{16\sigma\quad k_{w}}\left( \frac{B_{R}^{2}A_{m}l_{m}}{\mu_{r}} \right)}} \end{matrix} & {{Eq}.\quad 36} \end{matrix}$

The term A_(m)l_(m) represents the volume of PM material which, aside from gap radius r_(g), is the only dimensional variable pertinent to torque production.

Relative permeability μ_(r) of the PM material is often not found in the technical literature. Typically the “energy product” of PM material is published which is actually the “energy density” in kJ/meter³ which may be used in Eq. 27 by rearranging formula for energy density. Let: $\begin{matrix} {{E_{tot} = {\frac{B_{R}^{2}A_{m}l_{m}}{2\mu_{r}\mu_{o}} = {E_{\rho}A_{m}l_{m}}}};} & {{Eq}.\quad 37} \end{matrix}$ where:

-   -   E_(tot)=total PM magnetic energy stored in the volume of         A_(m)l_(m);     -   E_(ρ)=magnetic “energy product” of PM material, i.e., energy per         unit volume=kJ/meter³;         -   or: E_(ρ)=E_(tot)/A_(m)l_(m).             Then from Eq. 37: $\begin{matrix}             {\left( \frac{B_{R}^{2}A_{m}l_{m}}{\mu_{r}} \right) = {2\mu_{o}E_{\rho}A_{m}l_{m}}} & {{Eq}.\quad 38}             \end{matrix}$             Inserting Eq. 38 into Eq. 36: $\begin{matrix}             \begin{matrix}             {T = {\frac{{k_{F}\left( {P_{\Omega}/P_{S}} \right)}\omega\quad r_{g}^{2}}{16\sigma\quad k_{w}}2\mu_{o}E_{\rho}A_{m}l_{m}}} \\             {= {\frac{\mu_{o}E_{\rho}{k_{F}\left( {P_{\Omega}/P_{S}} \right)}\omega}{8\sigma\quad k_{w}}r_{g}^{2}A_{m}l_{m}}}             \end{matrix} & {{Eq}.\quad 39}             \end{matrix}$             Conversion Factors:

The following definitions and conversion constants may be used to convert Eq. 39 into English units: $\begin{matrix} {\mu_{o} = {\left( {1.26 \times 10^{- 6}} \right)\frac{{volt} - \sec}{{amp} - {meter}}}} & {{Eq}.\quad 40} \end{matrix}$ $\begin{matrix} {\sigma_{copper} = {\left( {1.7 \times 10^{- 8}} \right)\frac{{volt} - {meter}}{amp}}} & {{Eq}.\quad 41} \end{matrix}$ $\begin{matrix} \begin{matrix} {\omega = {\left( {2\pi\quad f} \right)\frac{1}{\sec}}} \\ {f = \frac{({rpm})}{60}} \end{matrix} & {{Eq}.\quad 42} \end{matrix}$ $\begin{matrix} \begin{matrix} {E_{\rho} = \frac{\left( {N - {meter}} \right)}{{meter}^{3}}} \\ {= \frac{Nm}{{meter}^{3}}} \\ {= \frac{{watt} - \sec}{{meter}^{3}}} \\ {= \frac{Joule}{{meter}^{3}}} \end{matrix} & {{Eq}.\quad 43} \end{matrix}$ $\begin{matrix} {(39.37)\quad\frac{in}{meter}} & {{Eq}.\quad 44} \end{matrix}$ $\begin{matrix} {({.7375})\quad\frac{{lb}.{ft}.}{Nm}\quad{or}\quad(1.356)\quad\frac{Nm}{{lb}.{ft}.}} & {{Eq}.\quad 45} \end{matrix}$ $\begin{matrix} {B = {\frac{{volt} - \sec}{{meter}^{2}} = {Telsa}}} & {{Eq}.\quad 46} \end{matrix}$ $\begin{matrix} {\frac{P_{\Omega}}{P_{S}} = {\left( {\frac{1}{E_{ff}} - 1} \right)\quad{from}\quad{{Eq}.\quad 15}}} & {{Eq}.\quad 47} \end{matrix}$ Rewriting Eq. 39 using Eqs. 40-44: $\begin{matrix} {T = {\frac{\begin{matrix} {{\left( {1.26 \times 10^{\quad^{- 6}}} \right)\quad{volt}} -} \\ {\sec\quad{k_{F}\left( {P_{\Omega}/P_{S}} \right)}2{\pi({rpm})}{E_{\rho}\left( {{Newton} - {meter}} \right)}} \end{matrix}}{\begin{matrix} {{8\left( {1.7 \times 10^{- 8}} \right)\quad\frac{{volt} - {meter}}{amp}k_{w}\quad{amp}} -} \\ {{{meter}\quad(60)\quad\frac{\sec}{\min}\quad\min} - {meter}^{3}} \end{matrix}}\frac{r_{g}^{2}{in}^{2}A_{m}{in}^{2}l_{m}{in}}{(39.37)^{3}\frac{{in}^{5}}{{meter}^{5}}}}} & {{Eq}.\quad 48} \end{matrix}$

All of the units cancel in Eq. 48, except for (Newton−meter) or Nm to use the common nomenclature. Thus torque in Eq. 48 is given in units of Newton-meters. Dropping the units in Eq. 48 for clarity gives: $\begin{matrix} {T_{Nm} = {\left\lbrack \frac{2{\pi\left( {1.26 \times 10^{- 6}} \right)}}{8\left( {1.7 \times 10^{- 8}} \right)(60)(39.37)^{5}} \right\rbrack\frac{k_{F}}{k_{w}}\left( {P_{\Omega}/P_{S}} \right)({rpm})E_{\rho}r_{g - {in}}^{2}A_{m - {in}}l_{m - {in}}}} & {{Eq}.\quad 49} \end{matrix}$ $\begin{matrix} {T_{Nm} = {\left( {1.025 \times 10^{- 8}} \right)\frac{k_{F}}{k_{w}}\left( {P_{\Omega}/P_{S}} \right)({rpm})E_{\rho}r_{g - {in}}^{2}A_{m - {in}}l_{m - {in}}}} & {{Eq}.\quad 50} \end{matrix}$ Using Eq. 45 to convert Eq. 50 to units of lb.ft.: $\begin{matrix} {T_{{lb},{ft}} = {\left( {7.56 \times 10^{- 9}} \right)\frac{k_{F}}{k_{w}}\left( {P_{\Omega}/P_{S}} \right)({rpm})E_{\rho}r_{g - {in}}^{2}A_{m - {in}}l_{m - {in}}}} & {{Eq}.\quad 51} \end{matrix}$ Substituting Eq. 47 into Eq. 51: $\begin{matrix} {T_{{lb}.{ft}.} = {\left( {7.56 \times 10^{- 9}} \right)\left( {\frac{1}{E_{ff}} - 1} \right)\frac{k_{F}}{k_{w}}({rpm})E_{\rho}r_{g - {in}}^{2}r_{g - {in}}^{2}A_{m - {in}}l_{m - {in}}}} & {{Eq}.\quad 52} \end{matrix}$

The dimensional terms in Eqs. 49-52 reveal the “5^(th) Power Rule” where: r²A r→X⁵, which is applicable to all electromagnetic machines including non-mechanical devices such as transformers. This rule states that for a given efficiency, frequency and machine shape, power increases as the 5^(th) power of any dimension. Therefore, doubling the size of the machine will result in a 32-fold increase in power while weight only increase as the cube. Thus “specific power,” power per unit weight, varies as the square of machine size which means larger machines are more power-dense than smaller machines.

Equivalent Amp-Turns of PM Material:

The equivalent amp-turns (ni)_(PM eq). that would be required to duplicate the mmf of PM material can be found starting with Eq. 37 where: $\begin{matrix} {E_{\rho} = \frac{B_{R}^{2}}{2\mu_{r}\mu_{o}}} & {{{Eq}.\quad 52}A} \end{matrix}$ Therefore Eq. 1 may be rewritten by multiplying top and bottom by “2”, “B_(R)” and substituting Eq. 52A: $\begin{matrix} \begin{matrix} {({ni})_{{PM}\quad{{eq}.}} = \frac{B_{R}l_{m}}{\mu_{r}\mu_{o}}} \\ {= {\left( \frac{B_{R}^{2}}{2\mu_{r}\mu_{o}} \right)\frac{2l_{m}}{B_{R}}}} \\ {= {2{l_{m}\left( \frac{E_{\rho}}{B_{R}} \right)}}} \end{matrix} & {{{Eq}.\quad 52}B} \end{matrix}$ Converting Eq. 52B to English units: $\begin{matrix} {({ni})_{{{PM}\quad{{eq}.}}\quad} = {{2{l_{m}\left( \frac{E_{\rho}}{B_{R}} \right)}} = \frac{\begin{matrix} {{2l_{m}} - {i\quad n^{1}} - 10^{3} - E_{\rho - {{kJ}/m^{3}}} - {volt}^{1} -} \\ {{amp} - \sec^{1} - {meter}^{2} - {meter}^{1}} \end{matrix}}{B_{R} - {volt}^{1} - \sec^{1} - {meter}^{3} - {(39.37){in}^{1}}}}} & {{{Eq}.\quad 52}C} \end{matrix}$ $\begin{matrix} {({ni})_{{PM} - {{eq}.}} = {(50.8)l_{m - {in}}\frac{E_{\rho - {{kJ}/m^{3}}}}{B_{R - {Tesla}}}}} & {{{Eq}.\quad 52}D} \end{matrix}$

It should be understood that amp-turns (ni)_(PM eq.) as given by Eq. 52D is not the motor drive current, but rather the equivalent ni that would be required for creating the machine's static magnetic field electromagnetically rather than with PM material.

Reluctance Matching:

Motor torque is shown by Eqs. 49-52 to be dependent on the product of the PM gap area A_(m) and the PM gap length l_(m) to give volume A_(m)l_(m) without regard to the particular shape of the PM cavity. Only the absolute volume A_(m)l_(m) is relevant to the magnitude of the torque developed. However, relative to machine efficiency, the shape of both volumes, the PM cavity and the rotor-stator gap volume, are of paramount importance, as shown by Eq. 32 under optimized conditions. It is the ratio of l_(m) to A_(m), i.e., PM gap shape, that determines rotor-stator gap shape at maximum efficiency, not their absolute values. Dividing both sides of Eq. 32 by absolute permeability μ_(o) gives: $\begin{matrix} {\left( \frac{l_{g}}{\mu_{o}A_{g}} \right)_{opt} = \left( \frac{l_{m}}{\mu_{r}\mu_{o}A_{m}} \right)} & {{Eq}.\quad 53} \end{matrix}$

These terms are simply expressions of the rotor-stator gap and PM gap reluctances as presented by Eq. 3. Therefore, Eq. 53 can be written as:

_(g-opt)=

_(m)  Eq. 54

Under optimal design conditions, the reluctances of both gaps are equal. This is because the electric circuit analogue, optimized for maximum power transfer, requires load resistance to be equal to the supply resistance, a condition known as “impedance matching”. The magnetic circuit counterpart to electrical resistance is termed “reluctance,” the resistance to flux flow. Eq. 54 specifies an equivalent “reluctance matching” in order to obtain maximum rotor-stator gap power from a given mmf, where the PM material is acting as the analogous “electrical power supply”.

Machine Flux Capacity:

To determine the rotor-stator gap flux density when the optimal design criteria are satisfied, Eq. 32A may be substituted into Eq. 12: $\begin{matrix} {B_{g - {opt}} = {\frac{B_{R}l_{m}}{A_{g}\left( {\frac{l_{m}}{A_{m}} + \frac{\mu_{r}l_{g - {opt}}}{A_{g}}} \right)} = {\frac{B_{R}l_{m}}{A_{g}\left( {\frac{l_{m}}{A_{m}} + \frac{l_{m}}{A_{m}}} \right)} = {\frac{1}{2}\left( \frac{A_{m}}{A_{g}} \right)B_{R}}}}} & {{Eq}.\quad 55} \end{matrix}$ Rearranging terms in Eq. 55: $\begin{matrix} {{\left( {A_{g}B_{g}} \right)_{opt} = {{\frac{1}{2}A_{m}B_{R}} = \theta_{opt}}}{{Or}\text{:}}} & {{Eq}.\quad 56} \\ {\theta_{opt} = {\frac{1}{2}\theta_{R}}} & {{Eq}.\quad 57} \end{matrix}$

According to Eqs. 56 and 57, circuit flux θ_(opt) under optimal design parameters is half the value of flux θ_(R) that would exist under closed circuit conditions, wherein the rating of B_(R) is obtained. This result is achieved because the presence of the rotor-stator gap with an optimized design doubles the closed circuit reluctance (Eq. 54). Hence circuit flux would predictably drop to half the closed-circuit value. Measurement of total machine flux corresponds to half the PM magnet rating when the optimized criteria have been incorporated into the design. However, core cross-sectional area is designed to carry total machine flux near saturation so that closed-circuit flux is not obtained because of flux capacity limits imposed by saturation.

Relative Permeability of PM Material:

The value of the PM material's relative permeability μ_(r), according to Eq. 33, may be used to implement an optimized design. This value can be found by rearranging the terms in Eq. 38: $\begin{matrix} {\mu_{r} = \frac{B_{R}^{2}}{2\mu_{o}E_{\rho}}} & {{Eq}.\quad 58} \end{matrix}$

Thus, with the rated technical data of “residual induction”, i.e., flux density B_(R), and “energy product” E_(ρ), it is possible to calculate the relative permeability μ_(r) for use in dimensional equations such as Eq. 33. Converting Eq. 58 to English units using Eqs. 40, 43, 46: $\begin{matrix} {u_{r} = \frac{B_{R}^{2} - {volt}^{2} - \sec^{2} - {amp}^{1} - {meter}^{1} - {meter}^{3}}{\begin{matrix} {2 - {meter}^{4} - \left( {1.26 \times 10^{- 6}} \right) - {volt}^{1} - \sec^{1} - E_{\rho} -} \\ \left( {10^{3} - {volt}^{1} - {amp}^{1} - \sec^{1}} \right) \end{matrix}}} & {{Eq}.\quad 59} \\ \begin{matrix} {\mu_{r} = {\frac{1}{2\left( {1.26 \times 10^{- 6}} \right)10^{3}}\frac{B_{R}^{2}}{E_{\rho}}}} \\ {{= {(397)\frac{B_{r}^{2}}{E_{\rho}}}};} \end{matrix} & {{Eq}.\quad 60} \end{matrix}$ where energy product E_(ρ) is in units of kJ/meter³.

In the case of a Neodymium-Iron-Boron (NIB) “super magnet”, where B_(R)=1.21 Tesla and E_(ρ)=303 kJ/m³, the relative permeability calculated from Eq. 60 is μ_(r)=1.92, a value that remains fairly constant among various NIB magnet grades. Solving Eq. 33 for l_(m): $\begin{matrix} {l_{m} = {{\mu_{r}\left( \frac{A_{m}}{A_{g}} \right)}l_{g - {opt}}}} & {{Eq}.\quad 61} \end{matrix}$

In the particular case of A_(m)=A_(g) and μ_(r)=1.92≈2, Eq. 61 results inn a PM cavity length l_(m) that is almost twice as long as the rotor-stator gap length l_(g) at A_(m)=A_(g).

Flux Density Inside of PM Material:

Substituting Eq. 6 into Eq. 11: $\begin{matrix} {{B_{R}l_{m}} = {A_{m}{B_{m}\left( {\frac{l_{m}}{A_{m}} + \frac{\mu_{r}l_{g}}{A_{g}}} \right)}}} & {{Eq}.\quad 62} \end{matrix}$ Solving Eq. 62 for B_(m): $\begin{matrix} {B_{m} = \frac{B_{R}l_{m}}{A_{m}\left( {\frac{l_{m}}{A_{m}} + \frac{\mu_{r}l_{g}}{A_{g}}} \right)}} & {{Eq}.\quad 63} \end{matrix}$ Inserting Eq. 32A into Eq. 63 to find B_(m) for an optimized design: $\begin{matrix} \begin{matrix} {B_{m - {opt}} = \frac{B_{R}l_{m}}{A_{m}\left( \frac{2l_{m}}{A_{m}} \right)}} \\ {= {\frac{1}{2}B_{R}}} \end{matrix} & {{Eq}.\quad 64} \end{matrix}$

Comparing Eq. 64 and Eq. 55 shows that B_(g-opt) and B_(m-opt) are not the same. The difference is due to the fact that the PM gap of B_(m-opt) is filled with PM material such that the circuit mmf (see Eq. 1) and cavity reluctance

_(m) both vary with a changing PM cavity volume. On the other hand, only the rotor-stator gap reluctance

_(g) varies with a changing rotor-stator gap volume while the circuit mmf remains constant.

To verify the veracity of Eqs. 55 and 64, solve Eq. 64 for B_(R) and substitute into Eq. 55: $\begin{matrix} \begin{matrix} {B_{g - {opt}} = {\frac{1}{2}\left( \frac{A_{m}}{A_{g}} \right)2B_{m - {opt}}}} \\ {= \frac{A_{m}B_{m - {opt}}}{A_{g}}} \end{matrix} & {{Eq}.\quad 65} \end{matrix}$  A _(g) B _(g-opt) =A _(m) B _(m-opt) or θ_(g)=φ_(m)  Eq. 66

Eq. 66 results in a constancy of flux that is independent of circuit location, as previously indicated by Eq. 6.

Optimal Volume of PM Cavity:

Using the foregoing equations, a machine may be designed based on optimized parameters and yielding maximum performance. In order to maximize machine materials utilization, the iron core should operate near saturation because the quantity of flux φ and current I are the primary criteria governing torque production, as revealed by Eq. 10. The total flux φ_(m) produced by the PM cavity is given from Eq. 64 as: $\begin{matrix} \begin{matrix} {\phi_{m} = {A_{m}B_{m}}} \\ {= {\frac{1}{2}A_{m}B_{R}}} \end{matrix} & {{Eq}.\quad 67} \end{matrix}$ The subscript “opt” has been dropped for simplification with the understanding that all parameters are hereafter considered optimal unless otherwise stated.

Flux is conducted axially through the rotor portion of the magnetic circuit at a flux density near saturation. Using Eq. 67 let: $\begin{matrix} \begin{matrix} {\phi_{m} = \phi_{{rotor}\quad}} \\ {= {A_{rotor}B_{sat}}} \\ {{= {\frac{1}{2}A_{m}B_{R}}};} \end{matrix} & {{Eq}.\quad 68} \end{matrix}$ where:

-   -   B_(sat)=saturation flux density         With reference to FIG. 2, rotor cross-sectional area is:         A _(rotor) =πr _(g) ²  eq. 69         Inserting Eq. 69 into Eq. 68: $\begin{matrix}         {{\pi\quad r_{g}^{2}B_{sat}} = {\frac{1}{2}A_{m}B_{R}}} & {{Eq}.\quad 70}         \end{matrix}$         Solving Eq. 70 for PM cavity area A_(m): $\begin{matrix}         {A_{m} = {2\pi\quad{r_{g}^{2}\left( \frac{B_{sat}}{B_{R}} \right)}}} & {{Eq}.\quad 71}         \end{matrix}$         According to Eq. 71, PM cavity area A_(m) varies only with gap         radius r_(g) because both B_(sat) and B_(R) are constants.         Solving Eq. 32A for PM cavity length l_(m) under optimized         conditions: $\begin{matrix}         {l_{m} = {A_{m}\left( {\mu_{r}\frac{l_{g}}{A_{g}}} \right)}} & {{Eq}.\quad 72}         \end{matrix}$         Inserting Eq. 71 into Eq. 72 $\begin{matrix}         {l_{m} = {2{\pi\mu}_{r}r_{g}^{2}\frac{l_{g}}{A_{g}}\left( \frac{B_{sat}}{B_{R}} \right)}} & {{Eq}.\quad 73}         \end{matrix}$

An increase of rotor-stator gap area A_(g) results in a decrease of PM cavity length l_(m), which seemingly means a reduction in PM material usage and lower cost. But according to Eq. 39, machine torque T is a function of gap radius r_(g) and PM volume A_(m)l_(m) so that nothing is gained in terms of reducing the quantity of PM material per unit torque by enlarging the rotor-stator gap area A_(g). Therefore, A_(g) is the minimum value necessary to conduct rotor flux. This means gap area A_(g) must be equal to rotor area A_(rotor) to maintain constant flux density near saturation B_(sat). Using Eq. 69: A _(g) =A _(rotor) =πr _(g) ²  Eq. 74 Inserting Eq. 74 into Eq. 73: $\begin{matrix} {l_{m} = {{2{\pi\backslash\mu_{r}}{r_{g}^{2}\backslash\frac{l_{g}}{\pi\backslash\quad r_{g}^{2}\backslash}}\left( \frac{B_{sat}}{B_{R}} \right)} = {2\mu_{r}{l_{g}\left( \frac{B_{sat}}{B_{R}} \right)}}}} & {{Eq}.\quad 75} \end{matrix}$ Dimensions of the optimum PM cavity volume are given by Eq. 71 and Eq. 75. Combining these equations gives the total PM cavity volume as: $\begin{matrix} {{A_{m}l_{m}} = {4{\pi\mu}_{r}r_{g}^{2}{l_{g}\left( \frac{B_{sat}}{B_{R}} \right)}^{2}}} & {{Eq}.\quad 76} \end{matrix}$ Torque Using Optimal PM Cavity Volume:

Machine torque produced with optimal PM volume may be found by substituting Eq. 76 into Eq. 39 for torque T: $\begin{matrix} \begin{matrix} {T = {\frac{\mu_{o}E_{\rho}{k_{F}\left( {P_{\Omega}/P_{S}} \right)}\omega}{8\sigma\quad k_{w\quad}}r_{g}^{2}4\pi\quad\mu_{r}r_{g}^{2}{l_{g}\left( \frac{B_{sat}}{B_{R}} \right)}^{2}}} \\ {= {\frac{\mu_{r}\mu_{o}\pi\quad E_{\rho}{k_{F}\left( {P_{\Omega}/P_{S}} \right)}\omega}{2\sigma\quad k_{w}}\left( \frac{B_{sat}}{B_{R}} \right)^{2}r_{g}^{4}l_{g}}} \end{matrix} & {{Eq}.\quad 77} \end{matrix}$ Rearranging terms in Eq. 77: $\begin{matrix} {T = {\left( \frac{\mu_{r}\mu_{o}\pi}{2\sigma} \right){E_{\rho}\left( {B_{sat}/B_{R}} \right)}^{2}\left( {k_{F}/k_{w}} \right)\left( {P_{\Omega}/P_{S}} \right)\omega\quad r_{g}^{4}l_{g}}} & {{Eq}.\quad 78} \end{matrix}$ Substituting Eq. 47 into Eq. 78: $\begin{matrix} {T = {\left( \frac{\mu_{r}\mu_{o}\pi}{2\sigma} \right){E_{\rho}\left( {B_{sat}/B_{R}} \right)}^{2}\left( {k_{F}/k_{w}} \right)\left( {{1/E_{ff}} - 1} \right)\omega\quad r_{g}^{4}l_{g}}} & {{Eq}.\quad 79} \end{matrix}$

At a specified efficiency E_(ff) and shaft frequency ω, machine torque is expressed in terms of two dimensions, rotor-stator gap radius r_(g) and rotor-stator gap length l_(g), retaining the “5^(th) power rule” as before. Machine torque is extremely sensitive to gap radius r_(g), varying as the fourth power of r_(g) for a given rotor-stator gap length l_(g). Rearranging terms in Eq. 79: $\begin{matrix} {T = {\frac{\pi}{2\sigma}\left( \frac{\mu_{r}\mu_{o}E_{\rho}}{B_{R}^{2}} \right){B_{sat}^{2}\left( {k_{F}/k_{w}} \right)}\left( {{1/E_{ff}} - 1} \right)\omega\quad r_{g}^{4}l_{g}}} & {{Eq}.\quad 80} \end{matrix}$ From Eq. 58: $\begin{matrix} {\frac{\mu_{r}\mu_{o}E_{\rho}}{B_{R}^{2}} = \frac{1}{2}} & {{Eq}.\quad 81} \end{matrix}$ Inserting Eq. 81 into Eq. 80: $\begin{matrix} {T = {\frac{\pi}{4\sigma}{B_{sat}^{2}\left( {k_{F}/k_{w}} \right)}\left( {{1/E_{ff}} - 1} \right)\omega\quad r_{g}^{4}l_{g}}} & {{Eq}.\quad 82} \end{matrix}$ Converting Torque Equations to English Units: Converting Eq. 82 to English units using Eqs. 41, 42, 46: $\begin{matrix} {T = \frac{\begin{matrix} {{amp} - {\pi\quad B_{sat}^{2}{volt}^{2{\backslash 1}}} -} \\ {{{\sec^{2{\backslash 1}}\left( {k_{F}/k_{w}} \right)}\left( {{1/E_{ff}} - 1} \right)2{\pi({rpm})}{\min^{1\backslash}{\left( {r_{g}^{4}l_{g}} \right){in}^{5\backslash}}}} - {meter}^{5\backslash}} \end{matrix}}{\begin{matrix} {{4\left( {1.7 \times 10^{- 8}} \right){volt}^{1\backslash}} - {meter}^{1\backslash} - {meter}^{4\backslash} -} \\ {\min^{1\backslash}{{- (60)}{\sec^{1\backslash}(39.37)}^{5}{in}^{5\backslash}}} \end{matrix}}} & {{Eq}.\quad 83} \end{matrix}$ The units remaining are: T=volt−amp−sec=watt−sec=Joules =Nm  Eq. 84 Consolidating terms in Eq. 83: $\begin{matrix} {T_{Nm} = {\left( \frac{2\pi^{2}}{4\left( {1.7 \times 10^{- 8}} \right)(60)(39.37)^{5}} \right){B_{sat}^{2}\left( {k_{F}/k_{w}} \right)}\left( {{1/E_{ff}} - 1} \right)({rpm})\left( {r_{g}^{4}l_{g}} \right)}} & {{Eq}.\quad 85} \end{matrix}$  T _(Nm)=(0.0512)B _(sat-Tesla) ²(k _(F) /k _(w))(1/E _(ff)−1)(rpm)r _(g-in) ⁴ l _(g-in)  Eq. 87 Using Eq. 45 to convert Eq. 86 to units of lb.ft.: T _(lb.ft.)=(0.0377)B _(sat-Tesla) ²(k _(F) /k _(w))(1/E _(ff)−1)(rpm)r _(g-in) ⁴ l _(g-in)  Eq. 87 As a cross-check for errors in Eq. 87, substitute Eq. 58 into Eq. 76 for PM cavity volume A_(m)l_(m): A m ⁢ l m = 4 ⁢ πμ r ⁢ r g 2 ⁢ l g ⁡ ( B sat B R ) 2 = 2 ⁢ π ⁢ R 2 ⁢ ( B sat 2 R 2 ) ⁢ = 2 ⁢ π μ o ⁢ E ρ ⁢ r g 2 ⁢ l g ⁢ B sat 2 Eq .   ⁢ 88 Converting Eq. 88 to English units: $\begin{matrix} \begin{matrix} {{A_{m}l_{m}} = \frac{\begin{matrix} {{2{\pi\left( {r_{g}^{2}l_{g}} \right)}{in}^{3}} - B_{sat}^{2} - {volt} - {\sec} -} \\ {{amp} - {meter} - {meter}} \end{matrix}}{\begin{matrix} {{\left( {1.26 \times 10^{- 6}} \right){volt}} - {\sec} -} \\ {{meter}{E_{\rho}\left( 10^{3} \right)}\left( {{{volt}\quad} - {amp} - {\sec}} \right)} \end{matrix}}} \\ {= {in}^{3}} \end{matrix} & {{Eq}.\quad 89} \end{matrix}$ $\begin{matrix} \begin{matrix} {\left( {A_{m}l_{m}} \right)_{{in}^{3}} = {\left( \frac{2\pi}{\left( {1.26 \times 10^{- 3}} \right)} \right)\frac{B_{{sat} - {Tesla}}^{2}r_{g - {in}}^{2}l_{g - {in}}}{E_{\rho}}}} \\ {{= {\left( {4.986 \times 10^{3}} \right)\frac{B_{{sat} - {Tesla}}^{2}r_{g - {in}}^{2}l_{g - {in}}}{E_{\rho}}}};} \end{matrix} & {{Eq}.\quad 90} \end{matrix}$ where energy product E_(ρ) is in units of kJ/meter³. Substituting Eq. 90 for A_(m)l_(m) into Eq. 51 for torque T: $\begin{matrix} {T_{{lb}.{ft}.} = {\left( {7.56 \times 10^{- 9}} \right)\left( {k_{F}/k_{w}} \right)\left( {{1/E_{ff}} - 1} \right){({rpm})\backslash E_{\rho}}r_{g\text{-}{in}}^{2}\frac{\left( {4.986 \times 10^{6}} \right)B_{sat}^{2}r_{g\text{-}{in}}^{2}l_{g\text{-}{in}}}{\backslash E_{\rho}}}} & {{Eq}.\quad 91} \end{matrix}$  T _(lb.ft.)=(0.0377)B _(sat-Tesla) ²(k _(F) /k _(w))(1/E _(ff)−1)(rpm)r _(g-in) ⁴ l _(g-in)=Eq. 87  Eq. 92 Drive Current:

To find drive current I required to produce torque under optimized design conditions, Eq. 14 may be expressed in terms of optimized quantities: $\begin{matrix} {T_{opt} = {\frac{1}{2\pi}I\frac{B_{R}l_{m}}{\left( {\frac{l_{m}}{A_{m}} + \left( \frac{\mu_{r}l_{g}}{A_{g}} \right)_{opt}} \right)}}} & {{Eq}.\quad 93} \end{matrix}$ Substituting Eq. 32A into Eq. 93: T = 1 2 ⁢ π ⁢ I ⁢ B R ⁢ m ⁢ A m 2 ⁢ m = 1 4 ⁢ π ⁢ I ⁢   ⁢ A m ⁢ B R Eq .   ⁢ 94 Inserting Eqs. 56 and 57 into Eq. 94: $\begin{matrix} \begin{matrix} {T_{opt} = {\frac{1}{2\pi}I\frac{\left( {A_{m}B_{R}} \right)}{2}}} \\ {= {\frac{1}{2\pi}I\frac{\theta_{R}}{2}}} \\ {= {\frac{1}{2\pi}I\quad\theta_{opt}}} \\ {= {\frac{1}{2\pi}{I\left( {A_{g}B_{g}} \right)}}} \end{matrix} & {{Eq}.\quad 95} \end{matrix}$ Inserting Eq. 71 for A_(m) into Eq. 95: T opt = 1 2 ⁢ ⁢ I ⁢   ⁢ R ⁢ ⁢ r g 2 ⁡ ( B sat R ) = 1 2 ⁢ I ⁢   ⁢ r g 2 ⁢ B sat Eq .   ⁢ 96 Solving Eq. 96 for drive current I while dropping subscript “opt” inasmuch as optimization conditions have already been specified following Eq. 67: $\begin{matrix} {I = \frac{2T}{r_{g}^{2}B_{sat}}} & {{Eq}.\quad 97} \end{matrix}$ Converting Eq. 97 to English units using Eqs. 43-46: $\begin{matrix} {I = \frac{\begin{matrix} {{2T} - {lb} - {ft} - (1.356) - {amp} -} \\ {{volt} - \sec - {meter}^{2} - {(39.37)^{2}{in}^{2}}} \end{matrix}}{{lb} - {ft} - r_{g}^{2} - {in}^{2} - B_{sat} - {volt} - \sec - {meter}^{2}}} & {{Eq}.\quad 98} \end{matrix}$ $\begin{matrix} {I = {\left( {4.203 \times 10^{3}} \right) = \frac{T_{{lb}.{ft}.}}{r_{g - {in}}^{2}B_{{sat} - {Tesla}}}}} & {{Eq}.\quad 99} \end{matrix}$ Induced Voltage:

Machine induced voltage v_(S), which is typically referred to as “back emf” in a motor and “forward emf” in a generator, excludes resistive voltage drop. Induced voltage v_(S) is related only to mechanical power P_(S).

From Eq. 22: P _(S) =Tω=Iv _(S)  Eq. 100 Solving Eq. 100 for machine voltage v_(S) and substituting Eq. 96 for torque T: v s = ω ⁢ T I = 1 2 ⁢ ω ⁢ ⁢ r g 2 ⁢ B sat = 1 2 ⁢ ω ⁢   ⁢ r g 2 ⁢ B sat Eq .   ⁢ 101

Arriving at Eq. 101 using a different approach will validate the preceding equations from which it was derived. The familiar equation for voltage v_(S) induced in a conductor of length w_(g) moving at velocity V through a magnetic field of flux density B_(g) is: v _(S) =Vw _(g) B _(g);  Eq. 102 where:

-   -   V=velocity of the conductor normal to the magnetic field; and     -   w_(g)=conductor length immersed in magnetic field=rotor/stator         gap axial length         Let tangential velocity V be expressed in terms of angular         frequency c:         V=ωr _(g)  Eq. 103         From Eq. 6 for constancy of flux throughout the entire machine         core, let:         φ_(g)=Φ_(rotor)  Eq. 104         Or:         A _(g) B _(g) =A _(rotor) B _(sat)  Eq. 105         Maximum core material utilization occurs when all of the iron is         operating near saturation so that:         B_(g)=B_(sat)         Therefore, substituting Eq. 106 into Eq. 105:         A_(g)=A_(rotor)         Expressing Eq. 107 in terms of rotor radius r_(g) and         rotor-stator gap length w_(g):         2πr _(g) w _(g) =πr _(g) ²  Eq. 108         Solving Eq. 108 for conductor length w_(g): $\begin{matrix}         {w_{g} = {\frac{1}{2}r_{g}}} & {{Eq}.\quad 109}         \end{matrix}$         Substituting Eqs. 103, 106 and 109 into Eq. 102: $\begin{matrix}         \begin{matrix}         {v_{S} = {\left( {\omega\quad r_{g}} \right)\frac{1}{2}r_{g}B_{sat}}} \\         {= {\frac{1}{2}\omega\quad r_{g}^{2}B_{sat}}} \\         {= {{Eq}.\quad 101}}         \end{matrix} & {{Eq}.\quad 110}         \end{matrix}$         Converting Eq. 110 to English units: $\begin{matrix}         {v_{S} = \frac{\begin{matrix}         {{2{\pi({RPM})}} - \min - r_{g}^{2} - {in}^{2} -} \\         {{meter}^{2} - B_{{sat} - {Telsa}} - {volt} - \sec}         \end{matrix}}{2 - \min - 60 - \sec - (39.37)^{2} - {in}^{2} - {meter}^{2}}} & {{Eq}.\quad 111}         \end{matrix}$  v _(S)=(3.378×10⁻⁵)(RPM)r _(g-in) ² B         _(sat-Tesla)  Eq. 112

It should be understood that machine voltage v_(S) as given by Eq. 112 is not the applied terminal voltage, but rather the voltage arising from mechanical power which excludes resistive voltage drop across the copper conductors.

Ratio of PM Cavity Area to Rotor-Stator Gap Area:

Substituting Eq. 106 into Eq. 55 gives the ratio of PM cavity cross-sectional area A_(m) relative to the rotor-stator gap area A_(g): $\begin{matrix} {{B_{g} = {B_{sat} = {\frac{1}{2}\frac{A_{m}}{A_{g}}B_{R}}}}{{Then}\text{:}}} & {{Eq}.\quad 113} \\ {\frac{A_{m}}{A_{g}} = {2\left( \frac{B_{sat}}{B_{R}} \right)}} & {{Eq}.\quad 114} \end{matrix}$

From the above equation it may be deduced that if B_(sat)=B_(R) then A_(g)=½A_(m). This results because machine flux is half the closed-circuit value due to double the reluctance relative to the closed-circuit reluctance at which B_(R) is rated. Also, the area ratio of Eq. 114 is constant irrespective of absolute machine size, being a function of magnetic properties only.

Ratio of PM Cavity Volume to Rotor-Stator Gap Volume:

From Eq. 74 let: A _(g) l _(g) =πr _(g) ² l _(g)  Eq. 115 Dividing Eq. 76 for A_(m) l_(m) by Eq. 115: A m ⁢ l m A g ⁢ l g = 4 ⁢ ⁢ μ r ⁢ g 2 ⁢ g ⁢ g 2 ⁢ g ⁢ ( B sat B R ) 2 = 4 ⁢ μ r ⁡ ( B sat B R ) 2 Eq .   ⁢ 116

According to Eq. 116, the ratio of the PM cavity vs. rotor-stator volume is constant regardless of absolute volume size. The volume ratio is strictly a function of magnetic properties independent of machine geometry or structural dimensions.

Examining Eq. 33 and Eq. 116 shows that the shape (Eq. 33) and size (Eq. 116) of the PM cavity corresponds to the shape and size of the rotor-stator gap volume that is chosen to satisfy a particular mechanical design, or vice-versa. In other words, both the shape and size of the rotor-stator gap volume are independent variables, while the PM cavity shape and size are dependent upon the rotor-stator geometry (or vice-versa) according to the dictates of Eqs. 33 and 116.

Summary of Equations

The dimensional parameters, given in “inch” units, of an optimized machine are summarized here with reference to FIG. 2. From Eq. 71: $\begin{matrix} \begin{matrix} {A_{m} = {2\pi\quad{r_{g}^{2}\left( \frac{B_{sat}}{B_{R}} \right)}}} \\ {{= {\pi\left( {r_{o}^{2} - r_{g}^{2}} \right)}};} \end{matrix} & {{Eq}.\quad 117} \end{matrix}$ where:

-   -   r_(o)=outside radius of PM cavity         Solving Eq. 113 for PM cavity outside radius r_(o):         $\begin{matrix}         {r_{o - {in}} = {r_{g - {in}}\left\lbrack {1 + {2\left( \frac{B_{sat}}{B_{R}} \right)}} \right\rbrack}^{1/2}} & {{Eq}.\quad 118}         \end{matrix}$         Rewriting Eq. 109: $\begin{matrix}         {w_{g - {in}} = {\frac{1}{2}r_{g - {in}}}} & {{Eq}.\quad 119}         \end{matrix}$         PM cavity length l_(m) is taken from Eq. 75: $\begin{matrix}         {l_{m - {in}} = {2\quad\mu_{r}{l_{g - {in}}\left( \frac{B_{sat}}{B_{R}} \right)}}} & {{Eq}.\quad 120}         \end{matrix}$

Because the PM relative permeability μ_(r) is not usually published, it may be calculated from Eq. 60, repeated here for convenience: $\begin{matrix} {\mu_{r} = {(397)\frac{B_{R}^{2}}{E_{\rho}}}} & {{Eq}.\quad 121} \end{matrix}$ Typically μ_(r)≈1.92 for neodymium-iron-boron magnets. (See the calculation following Eq. 60)

Substituting Eq. 121 into Eq. 120 as an alternative expression for PM cavity length l_(m) using published data of “residual induction” B_(R) and “energy product” E_(ρ): l m - in = 2 ⁢ ( 397 ) ⁢ ( B ⁢ R 1 E ρ ) ⁢ l g - in ⁡ ( B sat B ⁢ R ) = ( 794 ) ⁢ l g - in ⁡ ( B R ⁢ B sat E ρ ) ; Eq .   ⁢ 122 where flux density B is always in units of “Tesla” and “energy product” E_(p) is in units of kJ/meter³

If the PM cavity length l_(m) is known in advance, then the rotor-stator gap length l_(g) is derived from Eq. 122: $\begin{matrix} {l_{g - {in}} = {({.00126}){l_{m - {in}}\left( \frac{E_{\rho}}{B_{R}B_{sat}} \right)}}} & {{Eq}.\quad 123} \end{matrix}$ Total volume of PM material is found from Eq. 90: $\begin{matrix} {{{\left( {A_{m}l_{m}} \right)_{{in}^{3}} = {\left( {4.986 \times 10^{3}} \right)\frac{B_{{sat} - {Tesla}}^{2}r_{g - {in}}^{2}l_{g - {in}}}{E_{\rho - {{kJ}\text{/}m^{3}}}}}};}{where}{E_{\rho} = {{kJ}\text{/}m^{3}}}} & {{Eq}.\quad 124} \end{matrix}$ Let the weight W_(i-PM) of PM material be given as: $\begin{matrix} {\begin{matrix} {\left( W_{t - {PM}} \right)_{lbs} = {\left( {A_{m - {in}}l_{m - {in}}} \right)\rho_{PM}}} \\ {{= {\left( {A_{m - {in}}l_{m - {in}}} \right)({.272})\quad\frac{lb}{{in}^{3}}}};} \end{matrix}{{where}\text{:}}{\rho_{PM} = {{PM}\quad{material}\quad{density}\quad{of}\quad{approximately}\quad({.272})\quad{lb}\text{/}{in}^{\quad 3}}}} & {{Eq}.\quad 125} \end{matrix}$ Substituting Eq. 124 into Eq. 125: $\begin{matrix} {\left( W_{t - {PM}} \right)_{lbs} = {\left( {1.35 \times 10^{3}} \right)\frac{B_{{sat} - {Tesla}}^{2}r_{g - {in}}^{2}l_{g - {in}}}{E_{\rho - {{kJ}\text{/}m^{3}}}}}} & {{Eq}.\quad 126} \end{matrix}$ Repeated here for convenience are Eq. 92 and Eq. 52 for torque T: T _(lb.ft.)=(0.0377)B _(sat-Tesla) ²(k _(F) /k _(w))(1/E _(ff)−1)(rpm)r _(g-in) ⁴ l _(g-in)  Eq. 127 T _(lb.ft.)=(7.56×10⁻⁶)(k _(F) /k _(w))(1/E _(ff)−1)(rpm)E _(ρ-kJ/m) ₃ r _(g-in) ² A _(m-in) l _(m-in);  Eq. 128 where:

-   -   Energy product E_(ρ) has been changed to units of kJ/m³ in Eq.         124.

The DC homopolar motor/generator described herein specifies design parameters for achieving maximum torque at a given efficiency. Theoretical analysis reveals that optimum performance is obtained when the rotor-stator gap shape and size yields a total machine flux that is half of the closed-circuit flux content. (Closed-circuit flux is defined as the flux that would exist with a rotor-stator gap length of zero.) This ideal design condition is achieved when the machine's total magnetic circuit reluctance is divided equally between the reluctance of the PM cavity (filled with PM material) and the rotor-stator gap reluctance. The resulting “reluctance matching” is analogous to “impedance matching” in electrical circuits for maximum power transfer.

The optimized gap length of the DC homopolar motor/generator described herein is many times larger than that of a conventional homopolar machine. Although machine flux for a given amount of PM material is significantly less than found in standard practice, the attendant large copper volume permits high current at low dissipation to yield the highest torque theoretically possible from a given quantity of PM material.

This written description uses examples to disclose the invention, including the best mode, and also to enable a person skilled in the art to make and use the invention. The patentable scope of the invention may include other examples that occur to those skilled in the art. 

1. A DC homopolar machine, comprising: a stator that includes a stator magnetic core and a permanent magnet; a rotor magnetic core supported within the stator to rotate relative to the stator; a non-magnetic material within a gap between the rotor magnetic core and the stator magnetic core; and the stator, rotor and non-magnetic material forming a magnetic circuit having a total reluctance, the total reluctance of the magnetic circuit being provided by a first reluctance of the permanent magnet and a second reluctance of the gap of non-magnetic material, with the first reluctance being substantially equal to the second reluctance.
 2. The DC homopolar machine of claim 1, wherein the non-magnetic material includes copper.
 3. The DC homopolar machine of claim 1, wherein the non-magnetic material includes air.
 4. The DC homopolar machine of claim 1, wherein the non-magnetic material includes a rotor copper conductor coupled to the rotor magnetic core and a stator copper conductor coupled to the stator magnetic core.
 5. The DC homopolar machine of claim 4, wherein the non-magnetic material further includes an air gap between the rotor copper conductor and the stator copper conductor.
 6. The DC homopolar machine of claim 1, wherein the DC homopolar machine is a motor.
 7. The DC homopolar machine of claim 1, wherein the DC homopolar machine is a generator.
 8. The DC homopolar machine of claim 1, wherein the DC homopolar machine may operate as either a motor or a generator.
 9. The DC homopolar machine of claim 1, wherein a ratio of a length and a cross-sectional area of the permanent magnet is substantially equal to a ratio of a length and a cross-sectional area of the gap between the rotor magnetic core and the stator magnetic core multiplied by a relative permeability of the permanent magnet.
 10. The DC homopolar machine of claim 9, wherein the cross-sectional area of the permanent magnet is substantially equal to the cross-sectional area of the gap between the rotor magnetic core and the stator magnetic core multiplied by twice the core flux density divided by the residual induction flux density of the permanent magnet.
 11. The DC homopolar machine of claim 10, wherein the length of the permanent magnet is substantially equal to the length of the gap between the rotor magnetic core and the stator magnetic core multiplied by twice the relative permeability of the permanent magnet multiplied by the core flux density divided by the residual induction flux density of the permanent magnet.
 12. The DC homopolar machine of claim 1, wherein a cross-section area defined by the permanent magnet is about a minimum area necessary to conduct total machine flux at a flux density of half the residual induction flux density of the permanent magnet. 